The spin-only magnetic moment value of M³⁺ ion (in gaseous state) from the pairs Cr³⁺ / Cr²⁺, Mn³⁺ / Mn²⁺, Fe³⁺ / Fe²⁺ and Co³⁺ / Co²⁺ that has negative standard electrode potential, is ____________ B.M. [Nearest integer]

Solution

Chemistry ma’am · 3 steps

First find the one couple with a negative E°, then count unpaired electrons in its M³⁺ ion.

1. Couple	E° (V)	Sign
   Cr³⁺/Cr²⁺	-0.41	negative
   Mn³⁺/Mn²⁺	+1.57	positive
   Fe³⁺/Fe²⁺	+0.77	positive
   Co³⁺/Co²⁺	+1.97	positive

2. Why chromium. Cr³⁺ is 3d³, which half-fills the t_(2g) set, so it is very stable. It does not want to gain an electron, and that makes E° negative. The other three M³⁺ ions are easily reduced.
3. Count unpaired electrons. Cr is [Ar] 3d⁵ 4s¹. Removing three electrons gives Cr³⁺ as 3d³. In the gaseous ion each of the three electrons sits in a separate d orbital, so n = 3.

   μ = √n(n+2) = √(3 × 5) = √15 ≈ 3.87 B.M.

4. So. The spin-only magnetic moment of Cr³⁺ is 3.87 B.M., which rounds to 4.

The common slip: 5. Working with Cr²⁺ (3d⁴, n = 4) instead of the M³⁺ ion gives √24 ≈ 4.90, which rounds to 5. The question asks for the 3+ ion.

The idea to keep. Remember the M³⁺/M²⁺ values: only Cr is negative, because Cr³⁺ (t_(2g)³) is so stable. For μ, count unpaired electrons and use √n(n+2).

Answer from the NTA final answer key. Solution by PrepWiser.